Question

Bond order of ${O_2}$ , $O_2^ + $ , $O_2^ - $ and $O_2^{2 - }$ is in order ?
A. $O_2^ - < O_2^{2 - } < {O_2} < O_2^ + $
B. $O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $
C. $O_2^ + < O_2^{} < O_2^ - < O_2^{2 - }$
D. $O_2^{} < O_2^ + < O_2^ - < O_2^{2 - }$

Answer 1

Hint: We are provided with Oxygen molecules with different types of charge. We will first know what a bond order is. Then we will calculate the bond order of each molecule one by one. In the end we will arrange them in increasing order.

Formula Used:Bond Order: $\dfrac{1}{2} \times $$[$ Number of electron in A.MO. $]$$ - $ $[$ number of electrons in bonding molecular orbitals$]$

Complete step-by-step answer:
Step-1 All the molecules possess bond order. The bond order changes from one molecule to another molecule. So we should look at what the bond order is.
Step-2 Bond order is the difference between the number of electrons in bonding molecular orbitals and the number of electrons in the anti bonding molecular orbital.
Step-3 The bond order is directly proportional to the bond strength. If the bond order is higher than there is greater pull between two atoms. The bond order is inversely proportional to the shorter length of bond.
1. For ${O_2}$ :
${O_2}$ has a total 16 electrons in its orbitals. Molecular orbitals are completely full so they not participate in bonding
 Electronic configuration of ${O_2}$: ${({\sigma _{2s}})^2}\;{{\text{(}}{\sigma _{2s}}^*)^2}\;{{\text{(}}{\sigma _{2p}})^2}\;{{\text{(}}{\pi _{2P}})^4}\,{{\text{(}}{\pi _{2P}}^*)^2}$
So, bond order of ${O_2}$= $\dfrac{1}{2} \times \left[ {8 - 4} \right]$ , $ \Rightarrow $Bond order of ${O_2}$= $2$
For $O_2^ - $ : $O_2^ - $ has a total 17 electrons in its orbitals. Molecular orbitals are completely full so they not participate in bonding

2. Electronic configuration of $O_2^ - $ : ${({\sigma _{2s}})^2}\;{{\text{(}}{\sigma _{2s}}^*)^2}\;{{\text{(}}{\sigma _{2p}})^2}\;{{\text{(}}{\pi _{2P}})^4}\,{{\text{(}}{\pi _{2P}}^*)^3}$
Bond order for $O_2^ - $ = $\dfrac{1}{2} \times \left[ {8 - 5} \right]$
$ \Rightarrow $Bond order for $O_2^ - $ = 1.5
For $O_2^ + $ : $O_2^ + $ has a total 15 electrons in its orbitals.


3. Electronic configuration of $O_2^ + $ : ${({\sigma _{2s}})^2}\;{{\text{(}}{\sigma _{2s}}^*)^2}\;{{\text{(}}{\sigma _{2p}})^2}\;{{\text{(}}{\pi _{2P}})^4}\,{{\text{(}}{\pi _{2P}}^*)^1}$
Bond order for $O_2^ + $ = $\dfrac{1}{2} \times [8 - 3]$
$ \Rightarrow $Bond order for $O_2^ + $ = $2.5$

4. For $O_2^{2 - }$ : $O_2^{2 - }$ has a total 18 electrons in its orbitals .
Electronic configuration of $O_2^{2 - }$ : ${({\sigma _{2s}})^2}\;{{\text{(}}{\sigma _{2s}}^*)^2}\;{{\text{(}}{\sigma _{2p}})^2}\;{{\text{(}}{\pi _{2P}})^4}\,{{\text{(}}{\pi _{2P}}^*)^4}$
Bond order for $O_2^{2 - }$ = $\dfrac{1}{2} \times [8 - 6]$
$ \Rightarrow $Bond order for $O_2^{2 - }$ = $1$
So we get the increasing order of the bond order of the Oxygen molecule.
$O_2^{2 - } < O_2^ - < {O_2} < O_2^ + $ .

Note: Oxygen gas is a colorless, tasteless, odorless gas. It occurs in molecular nature. It is non metal. It is used for combustion. Still the pure oxygen does not burn. Till $1961$ oxygen was used as standard for weighing the elements. The excess oxygen is even harmful to the body.